作者：HurricaneCC | 来源：互联网 | 2024-10-24 18:35
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
return its level order traversal as:
提示：
此题要求逐行输出二叉树的节点数值，这里提供两种解法，第一种基于BFS，第二种基于DFS。
代码：
BFS：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vectorint>> levelOrder(TreeNode* root) {
vectorint>> res;
if (!root) return res;
queue q;
vector<int> v;
q.push(root);
q.push(NULL);
TreeNode *node;
while(!q.empty()) {
node = q.front();
q.pop();
if (node == NULL) {
res.push_back(v);
v.clear();
if (q.size()) q.push(NULL);
} else {
v.push_back(node->val);
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
}
return res;
}
};
DFS:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vectorint>> levelOrder(TreeNode* root) {
search(root, 0);
return res;
}
private:
void search(TreeNode* node, int depth) {
if (!node) return;
if (res.size() == depth) {
res.push_back(vector<int>());
}
res[depth].push_back(node->val);
search(node->left, depth + 1);
search(node->right, depth + 1);
}
vectorint>> res;
};
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